Problem: $\begin{aligned} &G(x)=\sqrt{3x} \\\\ &g(x)=G'(x) \end{aligned}$ $\int_{3}^{12} g(x)\,dx=$
Solution: $g$ is the derivative of $G$, which means $G$ is an antiderivative of $g$. Since we know the antiderivative of $g$, we can use the fundamental theorem of calculus: For every function $g$ and its antiderivative $G$, $\int_a^b g(x)\,dx=G(b)-G(a)$. $\begin{aligned} &\phantom{=}\int_{3}^{12} g(x)\,dx \\\\ &=G({12})-G({3}) \\\\ &=\sqrt{3{(12)}}-\sqrt{3{(3)}} \\\\ &=6-3 \\\\ &=3 \end{aligned}$ In conclusion, $\int_{3}^{12} g(x)\,dx=3$